Linear Algebra- independence and dependence Let $u$ and $v$ be non-zero vectors in $V$. Prove or disprove the following claim. 
$u$ and $v$ are linearly dependent $\implies$  $(u+v)$ and $(u-v)$ are linearly dependent.
Is the following proof correct?
Otherwise can someone answer the given question? Thanks in advance. :)
Here is my answer.
Since $u , v$ are two linearly dependent vectors, $au + bv = 0$ with $a,b \neq 0$
we can write,
$[(a+b)/2]\cdot(u+v) + [(a-b)/2] \cdot(u-v) = 0$
case 1 ($|a|=|b|$) :
if $a=b$ then $(a+b)/2 \neq 0 \implies (u+v)$ and $(u-v)$ are linearly dependent.
if $a= -b$  then $(a-b)/2 \neq 0\implies (u+v)$ and $(u-v)$ are linearly dependent.
case 2 ($|a| \neq|b|$) :
then $(a+b)/2, (a-b)/2 \neq 0\implies(u+v)$ and $(u-v)$ are linearly dependent.
So,
u and v are linearly dependent $\implies(u+v)$ and $(u-v)$ are linearly dependent.
 A: Your proof is correct. However, I don't see why you need to break it up into those 2 cases.
Like you said, $au + bv = 0 \Leftrightarrow (a, b)=(0,0)$.
Now, assume that $a(u+v) + b(u-v) = 0$, so $(a+b) u + (a-b) v = 0$. Hence, we know that $a+b = 0 $ and $a-b=0$, which gives that $2a=0, 2b=0$. (Assuming that your field does not have characteristic 2) hence $a=0, b=0$.
In the event that your field has characteristic 2, then $u+v = u-v$ so these vectors are not linearly independent.

Edit: OP wanted a direct proof along the lines he started, and I pointed out that he doesn't need to split it out.
Assume that $u, v$ are not linearly independent. Hence, there exists $(a, b) \neq (0,0)$ such that $au + bv = 0$.
This implies that $ \frac {a+b}{2} (u+v) + \frac {a-b}{2} (u-v) = au +bv = 0$ (assuming that the field does not have characteristic 2). For this to actually show that $(u+v), (u-v)$ are not linearly independent, we have to show that $(\frac {a+b}{2}, \frac {a-b}{2}) \neq (0,0)$. This is true because $ (\frac {a+b}{2}, \frac {a-b}{2}) = (0,0) \Leftrightarrow (a, b) = (0,0)$.
A: since {u,v} are linearly independent, the set {u+v,u-v} is LI if and only if the determinant formed by the coefficients that is 
        1  1
        1  -1 
is non zero
